3.00 - Review of Trigonometry

University:
High School
Course:
High School Physics
Academic year:

2024
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412

Pages:

15
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Trigonometry 3.00: Right-angle trig, Sine law, Cosine law, Reciprocal Trig Ratios The Trigonometric Ratios O SINE O COSINE O sin 𝜃 = 𝑜𝑝𝑝 ℎ𝑦𝑝 O cos 𝜃 = 𝑎𝑑𝑗 ℎ𝑦𝑝 O tan 𝜃 = 𝑜𝑝𝑝 𝑎𝑑𝑗 = 𝑂 𝐻 = 𝐴 𝐻 = 𝑂 𝐴 B O TANGENT a C b A Sine Law 𝑎 𝑏 𝑐 = = sin 𝐴 sin 𝐵 sin 𝐶 Cosine Law 𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos 𝐶 𝑐 2 − 𝑎2 − 𝑏2 cos 𝐶 = −2𝑎𝑏 Area 𝑏ℎ 𝐴= 2 𝐴= 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) 𝑠= 𝑎+𝑏+𝑐 2 𝑎𝑏 sin 𝐶 𝐴= 2 semi-perimeter The Trigonometric Ratios ANGLE A O O sin 𝐴 = 𝑎 𝑐 ANGLE B O sin 𝐵 = 𝑏 𝑐 O cos 𝐴 = 𝑏 𝑐 O cos 𝐵 = 𝑎 𝑐 O tan 𝐴 = 𝑎 𝑏 O tan 𝐵 = 𝑏 𝑎 B a C b A Complementary Angles with Sin & Cos Since ∆𝐴𝐵𝐶 is a right triangle: O ∠𝐴 + ∠𝐵 + ∠𝐶 = 180 O ∠𝐶 = 90° ∠𝐴 + ∠𝐵 + 90 = 180 ∠𝐵 = 90 − ∠𝐴 O B O ∠𝐴 C b AND ∠𝐵 are complementary angles a A 30-60-90 Triangle If ∆𝐴𝐵𝐶 is a right triangle and ∠𝐴 = 30 O the side opposite the 30° is equal to half of the hypotenuse 1 𝑎 = ℎ𝑦𝑝 2 B a C b A 30-60-90 Triangle B 60° Why? 30° x A x x/2 60° 60° x C Tangent Identity sin 𝐴 cos 𝐴 𝑎 =𝑐 𝑏 𝑐 𝑎 𝑐 = × 𝑐 𝑏 𝑎 = 𝑏 = tan 𝐴 B a C b A sin 𝑨 tan 𝑨 = cos 𝑨 The Reciprocal Ratios O COSECANT O SECANT O csc θ = 1 sin 𝜃 = 𝐻 𝑂 O sec θ = 1 cos 𝜃 = 𝐻 𝐴 O cot θ = 1 tan 𝜃 = 𝐴 𝑂 B O COTANGENT a C b A The Reciprocal Ratios ANGLE A O ANGLE B O sec 𝐴 = 𝑐 𝑏 O sec 𝐵 = 𝑐 𝑎 O csc 𝐴 = 𝑐 𝑎 O csc 𝐵 = 𝑐 𝑏 O cot 𝐴 = 𝑏 𝑎 O cot 𝐵 = 𝑎 𝑏 B a C b A Cotangent Identity csc 𝐴 sec 𝐴 𝑐 =𝑎 𝑐 𝑏 𝑐 𝑏 = × 𝑎 𝑐 B 𝑏 = 𝑎 = cot 𝐴 a C b A csc 𝑨 cot 𝑨 = sec 𝑨 Ex 1: find six exact trig ratios and solve for angle 122 − 92 =3 7 12 ϴ 9 𝑂 3 7 7 sin 𝜃 = = = 𝐻 12 4 cos 𝜃 = 𝐴 9 3 = = 𝐻 12 4 𝑂 3 7 7 tan 𝜃 = = = 𝐴 9 3 3 𝜃 = 𝑐𝑜𝑠 −1 ( ) ≈ 41.4° 4 csc 𝜃 = 𝐻 4 4 7 = = 𝑂 7 7 sec 𝜃 = 𝐻 4 = 𝐴 3 𝐴 3 3 7 cot 𝜃 = = = 𝑂 7 7 HW Worksheet on Teams: Right Angle Trig Evaluating Ratios.pdf